Solve the quadratic equations by factoring.

1. 5x^2 = 3x + 2

2. x^2 - 4x + 3 = 0

Help on these two problems above would be very appreciated, thanks!



Answer :

Solving by factoring in steps shown below

Question 1

  • 5x² = 3x + 2
  • 5x² - 3x - 2 = 0
  • 5x² - 5x + 2x - 2 = 0
  • 5x(x - 1) + 2(x - 1) = 0
  • (5x + 2)(x - 1) = 0
  • 5x + 2 = 0 and x - 1 = 0
  • 5x = - 2 and x = 1
  • x = - 2/5 and x = 1

Question 2

  • x² - 4x + 3 = 0
  • x² - x - 3x + 3 = 0
  • x(x - 1) - 3(x - 1) = 0
  • (x - 3)(x - 1) = 0
  • x - 3  = 0 and x - 1 = 0
  • x = 3 and x = 1

Answer:

[tex]\textsf{1. \quad $x=1, \quad x=-\dfrac{2}{5}$}[/tex]

[tex]\textsf{2. \quad $x=1, \quad x=3$}[/tex]

Step-by-step explanation:

Solving quadratic equations by factoring

  • To factor a quadratic in the form [tex]ax^2+bx+c[/tex] , find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers.
  • Factor the first two terms and the last two terms separately.
  • Factor out the common term.
  • Solve for x by applying the zero-product property.

Question 1

Given equation:

[tex]5x^2=3x+2[/tex]

Subtract (3x + 2) from both sides:

[tex]\implies 5x^2-(3x+2)=3x+2-(3x+2)[/tex]

[tex]\implies 5x^2-3x-2=0[/tex]

Find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers:

[tex]\implies ac=5 \cdot -2=-10[/tex]

[tex]\implies b=-3[/tex]

Therefore, the two numbers are -5 and 2.

Rewrite [tex]b[/tex] as the sum of the two numbers:

[tex]\implies 5x^2-5x+2x-2=0[/tex]

Factor the first two terms and the last two terms separately:

[tex]\implies 5x(x-1)+2(x-1)=0[/tex]

Factor out the common term (x - 1):

[tex]\implies (5x+2)(x-1)=0[/tex]

Apply the zero-product property:

[tex]\implies 5x+2=0 \implies x=-\dfrac{2}{5}[/tex]

[tex]\implies x-1=0 \implies x=1[/tex]

Therefore, the solutions are:

[tex]\boxed{x=1, \quad x=-\dfrac{2}{5}}[/tex]

Question 2

Given equation:

[tex]\implies x^2-4x+3=0[/tex]

Find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers:

[tex]\implies ac=1 \cdot 3=3[/tex]

[tex]\implies b=-4[/tex]

Therefore, the two numbers are -3 and -1.

Rewrite [tex]b[/tex] as the sum of the two numbers:

[tex]\implies x^2-3x-x+3=0[/tex]

Factor the first two terms and the last two terms separately:

[tex]\implies x(x-3)-1(x-3)=0[/tex]

Factor out the common term (x - 3):

[tex]\implies (x-1)(x-3)=0[/tex]

Apply the zero-product property:

[tex]\implies x-1=0 \implies x=1[/tex]

[tex]\implies x-3=0 \implies x=3[/tex]

Therefore, the solutions are:

[tex]\boxed{x=1, \quad x=3}[/tex]